# $L^2$ convergence of holomorphic functions

I am trying to solve a problem from this website that asks the following:

Prove that, for a sequence of holomorphic functions on a compact set, convergence in $L^2$ implies uniform convergence.

I am struggling with how to relate the $L^2$ norm to the fact that the functions are holomorphic. There is also a related question, which asks if the space of square-integrable holomorphic functions on a bounded region $U\subseteq\mathbb C$ is a Banach space. This seems like it would follow from showing that a sequence of functions that is "$L^2$-Cauchy" is uniformly Cauchy, but I'm not sure.

\begin{align*} |f_{n}(w)-f_{m}(w)|&=\left|\dfrac{1}{2\pi i}\displaystyle\int_{\gamma}\dfrac{f_{n}(z)-f_{m}(z)}{z-w}dz\right|\\ &\leq\dfrac{1}{2\pi}\left(\displaystyle\int_{\gamma}|f_{n}(z)-f_{m}(z)|^{2}|dz|\right)^{1/2}\left(\displaystyle\int_{\gamma}\dfrac{1}{|z-w|^{2}}|dz|\right)^{1/2}, \end{align*} where we can choose $\gamma$ to be the boundary of a closed ball $B_{M}(0)$ such that the compact $K$ in question satisfies $K\subseteq B_{M/2}(0)$, then the term $\displaystyle\int_{\gamma}\dfrac{1}{|z-w|^{2}}|dz|$ can be controlled uniformly on $w\in K$.
• We only know that the functions are defined on $K$. I think it saying $f\in L^2$ means that $\int_K |f(z)|^2\,dA<\infty$. – justintomb Mar 7 '18 at 3:07
• I think it means that it is defined on an open set containing the compact set $K$. – justintomb Mar 7 '18 at 3:09
• Then it could be very hard to solve. My philosophy is that, use a closed curve that encircled the whole compact set $K$, I have met this question with the assumption that $f_{n}$ being defined on a closed ball. – user284331 Mar 7 '18 at 3:12
Let $f$ be a function that is holomorphic on an open set $U$ containing a compact set $K$. Let $\delta>0$ be the distance from $\mathbb C\setminus U$ to $K$. If $z_0\in K$ and $r<\delta$, then by the Mean Value Property and the Cauchy-Schwarz inequality, $$|f(z_0)|=\left|\frac{1}{2\pi}\int_0^{2\pi} f(z_0+re^{i\theta})\,d\theta\right|\leq\frac{1}{2\pi}\int_0^{2\pi}|f(z_0+re^{i\theta})\cdot 1|\,d\theta$$ $$\leq\left(\frac{1}{2\pi}\int_0^{2\pi}|f(z_0+re^{i\theta})|^2\,d\theta\right)^{1/2}.$$ Thus, $$\frac{\delta}{2}|f(z_0)|^2\leq\int_{\delta/2}^\delta\frac{1}{2\pi}\int_0^{2\pi}|f(z_0+re^{i\theta})|^2\,d\theta\,dr\leq\frac{1}{2\pi}||f||_2^2.$$ Now suppose $(f_n)$ is a sequence of functions that are holomorphic on $U$. It follows from the above estimate that $L^2$ convergence of $f_n$ implies uniform convergence.